Horizontal Force on the Jounama Dam Radial Gate Pivots: 04-BS-7, May 2015, Q3

Question 3: Horizontal Force on the Jounama Dam Radial Gate Pivots (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given (from the attachment drawing, Problem 3.13 Jounama Dam):

Quantity	Value
Full Supply Level (F.S.L.)	RL 392.58 m
Spillway sill / gate bottom	RL 381.00 m
Radial gate size (width × height)	14.73 m × 11.58 m
Water density	1000 kg/m³

Find. The total horizontal hydrostatic force each gate's pivots must carry at full supply level.

Fig. Q3 — the gate submerged height h = FSL − sill = 392.58 − 381.00 = 11.58 m matches the printed gate height exactly, confirming the sill is the bottom of the wetted gate.

Approach. A radial (tainter) gate is a circular-arc surface, so every elemental pressure force points straight at the trunnion (pivot) and the net force also passes through it. The horizontal component of that resultant depends only on the gate's vertical projection — a flat rectangle of the gate's own height and width — so it can be found exactly as for a plane vertical gate, without needing the arc radius at all.

Confirm the submerged height from the two reference levels: $$h = 392.58 - 381.00 = 11.58\ \text{m}$$
Centroid depth of the vertical projection (rectangle, top at the free surface): $$\bar{h} = \frac{h}{2} = 5.79\ \text{m}$$
Horizontal force on the projected area A = w×h: $$F_H = \rho g \bar{h} A = \rho g \left(\frac{h}{2}\right)(w\,h) = 1000(9.81)(5.79)(14.73\times11.58)$$ $$F_H = \boxed{9689\ \text{kN} = 9.69\ \text{MN per gate}}$$
Depth of the line of action (for reference — this is where the pivot must sit for zero net moment): $$y_{cp} = \frac{2h}{3} = \frac{2(11.58)}{3} = 7.72\ \text{m below the F.S.L.}$$

Quantity	Result
Submerged gate height h	11.58 m
Horizontal force per gate, F_H	9.69 MN
Depth of resultant below FSL	7.72 m

Topic: Hydrostatic force on a curved (radial/tainter) gate.

Key relations: F_H = ρḡhA on the vertical projection; y_cp = 2h/3 for a surface-piercing rectangle.

Why this works. The horizontal component of pressure force on ANY surface equals the force on that surface's vertical projection — a direct consequence of resolving each pressure element (which acts normal to the local surface) into horizontal and vertical components and integrating. The gate's curvature only matters for the vertical component and for where the resultant passes (through the trunnion, by design, which is why radial gates need almost no force to rotate open).

Common pitfall. Trying to integrate around the curved arc directly, or assuming the trunnion location is needed to find F_H — it isn't; only the vertical projection matters for the horizontal component.

Source: White, Fluid Mechanics, Ch. 2 (forces on curved surfaces).

Question 3 of 13: Horizontal Force on the Jounama Dam Radial Gate Pivots

Question 3: Horizontal Force on the Jounama Dam Radial Gate Pivots (5 marks)